/*
To the Max
Time Limit: 1 Second      Memory Limit: 32768 KB

Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].


Output

Output the sum of the maximal sub-rectangle.


Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 
*/


/*
 Result:

 1731154 2008-12-22 20:51:20  Accepted 1074	C++ 440 8368 Jun
 */
#include <iostream>
#include <vector>

using namespace std;

typedef vector<vector<int> >  MATRIX;
typedef vector<MATRIX> CUBE;


int N=0;

inline int calc(int lr,int lc, int rr,int rc,CUBE& rv, CUBE& cv) {
	if (lr>=rr || lc>=rc) return 0;

	int ret=0;
	if (rr-lr>rc-lc)
		for (int i=lc;i<=rc;i++) 
			ret += cv[lr][i][rr];
	else 
		for (int i=lr;i<=rr;i++) 
			ret += rv[i][lc][rc];

	return ret;
}


int main(){
	int max=-127;
	int t,count=0;

	cin>>N;

	MATRIX M(N,vector<int>(N));
	CUBE RV(N,vector<vector<int> >(N,vector<int>(N)));
	CUBE CV(N,vector<vector<int> >(N,vector<int>(N)));

	while (cin>>t) {
		if (max<t) max = t;
		M[count/N][count%N] = t;
		++ count;
	}

	for (int i=0;i<N;i++) 
		for (int j=0;j<N;j++) {
				RV[i][j][j] = M[i][j];
				CV[i][j][i] = M[i][j];
			}

	for (int i=0;i<N;i++) 
		for (int j=0;j<N;j++) 
			for (int k=j+1;k<N;k++) 
				RV[i][j][k] = RV[i][j][k-1]+M[i][k];

	for (int j=0;j<N;j++) 
		for (int i=0;i<N;i++) 
			for (int k=i+1;k<N;k++)
				CV[i][j][k] = CV[i][j][k-1]+M[k][j];

	M.clear();

	for (int i=0;i<N;i++) 
		for (int j=0;j<N;j++) 
			for (int x=i+1;x<N;x++) 
				for (int y=j+1;y<N;y++) {
					if ( RV[i][j][y]<=0 || RV[x][j][y]<=0 ||
						 CV[i][j][x]<=0 || CV[i][y][x]<= 0)
						continue;

					t=calc(i,j,x,y,RV,CV);
					if (t>max) 
						max = t;
				}
	
	cout<<max<<endl;
}
